Refer to the following figure below showing the hybridized (green) and unhybridized (blue) atomic orbitals of a diatomic molecule. What is the hybridization of the ...

What is the hybridization of the indicated atom in each of the following molecules?

Hey, we're just gonna be looking out hybridization. We're gonna be identifying the hybridization of all of our circled centers. So just as a rule of thumb here, if we have single bonds, it's sp three. We have double bonds, not becomes sp two. And then if we have triple bonds, that is S P. So using this information, we can now identify all of our centers. So our first example we haven't sp to hybridize center That is a carpet in our next example off carbon I'll Structure weekend haven't SP two hybrid I sent moving on Our oxygen is now sp three We only have single bonds. Next we've got a nitrogen it in a triple bond and that is S p hybridization, followed by another nitrogen example. However, this time it's sp two hybridized and lastly, we just haven't sp three hybridized oxygen again

This is the answer to Chapter one problem Number 29 from the Smith Organic Chemistry Textbook on this problem asks us Thio, identify. Um the hybridization around indicated atoms in each of these three molecules, Um and so sort of ah, easy way Thio. To do that is related to the way that we were identifying geometry around particular Adams earlier in this chapter So we can count the groups around each Adam, um, and then correlate that with a particular orbital hybridization. Um and so, in a way, this first carbon has four groups around it, so it has three hydrogen tze and then another carbon and so four groups eyes going to be S p three. So it's tetra hydro geometry. When we have tetra hydro geometry, we have sp three orbital's, um and then looking at the other carbon that were asked about in a um it has two groups around it. Ah, carbon and hydrogen. Um, also, as I've said before, any molecule that's part of a triple bond or any Adam that's part of a triple bond in a molecule. I's gonna have linear geometry. So it's also true that it's gonna have SP hybridization. Um and so this carbon does have linear geometry and it has sp hybridization, um and so will follow the strategy for the other two molecules in this problem. So and be the carbon that were asked about has three groups around it. Um, so it's gonna have tried anal plainer geometry on DSO. It's gonna be SP too hybridized, Um, something that you It would be good to learn, though. That, you may want to know is when a carbon is making a double bond to something, it's sp two hybridized on dso Any time that you see a carbon with a double bond, it's gonna be SP two as it is here. So then this nitrogen has three groups around it, so it has this carbon to which it's double bound. It has a carbon on the other side, and it has a loan pair. So it's also going to be SP two hybridized. But again, it's also part of a double bond on that can tell us just by looking at it that it's gonna be sp two hybridized, um and so lastly for C um so, for see the first carbon that were asked about has three groups around it. It has to Hodgins, and then it has a carbon. I mean, it's also participating in this double bond. Um, So again, just by looking, we should be able to say S p to, um now, this second carbon is gonna be a little different. So the second carbon is participating in a double bond, but it's also participating in a second double bomb. So we do need to be mindful that, um And so here it might be best thio revert to counting the groups, as the textbook likes to talk about. So this carbon has one carbon to one side and one carbon to the other for a total of two groups. And so that tells us that it's going to be SP hybridized on. And so it is, has a linear geometry is sort of locked into place by these double bonds to two other carbons. Um, and that's the answer to Chapter one. Problem number 29

Why is the nitrogen of acid Hamid S P to hybridized and not as P three as it looks? Well, let's go ahead and draw. The orbital diagram is the problem suggests, and maybe that'LL help offer us a little bit more information. So I've drawn kind of in a flat mode. Of course, thes hydrogen is he wouldn't actually draw as a wedge or a dash, and you wouldn't draw. This methyl group is a tache, either, but it might help you to visualize this is plainer and go ahead and draw in the orbital's, in which oxygen and carbon are overlapping to form of high bond. Well, pi bonds are always made up of P orbital's. Therefore, we're talking about this p orbital overlap. The other important note about this compound is we can draw a resonant structure, negative charge on oxygen that would also put a positive charge on this nitrogen. And of course, it still has its high traditions attached to it. Okay, so in the residence form a residence form, there's a double bond between nitrogen and this carbon. We know that resident structures are really just ah, hybrid or compounds are really just a hybrid of the residence forms. So even though this isn't a great resonance structure, it does exist too. In a hickory, it's a minor contributor to the overall resident structure and we might want to draw a hybrid that looks something like this. All right. Therefore, since you can draw this orbital overlap, that means that the people there is a P orbital in the residence structure that is overlapping with the P orbital on carbon here. So nitrogen has tohave a P orbital strap it again, and that orbital is overlapping with carbons P orbital. So this is kind of what we're drawing in our overall resonance hybrid. Therefore, nitrogen is using a P orbital toe. Hold this loan pair of electrons because then it could be used to overlap with carbons p orbital. So this is the kind of orbital overlap you want to draw. To explain why this loan pair is in a p orbital and therefore why this nitrogen must be as p two hybridized because it can't be S p three hybridized because it's using a pew orbital to overlap and form a pie bond. One has taken away their four. There's only two piece and an s left over to hybridize to form signal bonds

Came. So the arrangement off the atoms is already shown on the white board on Now we know that carbon and hydrogen can only form one single bound. So what we can do now is to connect. All of the hydrogen is through the carbons through single bombs. We also know that between car carbon carbons, ears should be at least one bones between them. So we can also connect carbons with, um, we single bones right now. Okay, let's start with cover number one. We can see that it has already being connected to four atoms with four single bones. So we know the, um, hybridization for these carbon welby sp saree Okay. For the second carbon from the left, it is also have ah, four single bonds to it. So it should also have the SP three hybridization. Hey, for cover number three, we see that the connection towards the carbon is not complete. However, we also know that we cannot add any more bonds to thes single bound because otherwise the cover number to Welby have well have additional bounds. So we have to only add carbon bond through thes bonding. So to make we know that each carbon has to be connected with £4. So two more additional pounds will be added between no cover number two and cover and cover. Number three uncovered. Number four. So Okay, well, for the ease off explanation, let me number out each individual. Carbon five on six. Okay. For so cover number three, uncover number four should have a triple bomb on. Because all that triple bond, there's only one sigma bond on three pie bonds. So cover number three would only have as Pito hybridization. Hey on. Same thing with cover number four, which has one triple bound on one single bound, which should also only have SP to SP hybridization on for these carbon cover number five. It's right now. It is only has it only has three single bones. So to make the bonds complete, we need to add additional bound between covering number five. Uncover number six. So right now, Carver number five has one double bond on two single bonds. Souls the hybridization Welby SP two on the same thing with cover number six, it has one double bound to single bones, which is a total of three regions off electron density. So it will also has SP two hybridization